$\text{Bài 2:}$
$\text{a.}$
$\text{$n_{Fe_{2}O_{3}}$ = 16/160 = 0.1 mol}$
$\textit{$Fe_{2}O_{3}$ + 6HCl → $2FeCl_{3}$ + $3H_{2}O$}$
$\text{⇒ $n_{HCl}$ = 6$n_{Fe_{2}O_{3}}$ = 0.1*6 = 0.6 mol ⇒ $C_{M(HCl)}$ = 0.6/(100/1000) = 6 M}$
$\text{b.}$
$\text{$V_{H_{2}}$ = 0 lít (phản ứng không sinh ra khí $H_{2}$)}$
$\text{c. $n_{FeCl_{3}}$ = 2$n_{Fe_{2}O_{3}}$ = 2*0.1 = 0.2 mol ⇒ $m_{FeCl_{3}}$ = 0.2*162.5 = 32.5 g}$
$\text{Bài 3:}$
$\text{a.}$
$\text{$n_{H_{2}}$ = 3.36/22.4 = 0.15 mol}$
$\textit{$2Al_{ }$ + 6HCl → $2AlCl_{3}$ + $3H_{2}$↑}$
$\text{$n_{Al}$ = 2/3$n_{H_{2}}$ = 2/3*0.15 mol ⇒ $m_{Al}$ = 0.1*27 =2.7 mol}$
$\text{b. $n_{HCl}$ = 2$n_{H_{2}}$ = 2*0.15 = 0.3 mol ⇒ $V_{dd(HCl)}$ = 0.3/2 = 0.15 lít}$
$\text{Bài 4:}$
$\text{a.}$
$\text{$n_{CuSO_{4}}$ = 1*(100/1000) = 0.1 mol}$
$\textit{$2NaOH_{ }$ + $CuSO_{4}$ → $Na_{2}SO_{4}$ + $Cu(OH)_{2}$↓}$
$\text{$n_{NaOH}$ = 2$n_{CuSO_{4}}$ = 2*0.1 = 0.2 mol ⇒ $V_{dd(NaOH)}$ = 0.2/2 = 0.1 lít}$
$\text{b. $n_{Na_{2}SO_{4}}$ = $n_{Cu(OH)_{2}}$ = 0.1 mol ⇒ $m_{Na_{2}SO_{4}}$ = 0.1*142 = 14.2 g}$
